Question

Suppose four distinct positive numbers $a_{1}, a_{2}, a_{3}$ and $a_{4}$ are in $G.P$. Let $b_{1}=a_{1}, b_{2}=b_{1}+a_{2}, b_{3}=b_{2}+a_{3}$ and $b_{4}=b_{3}+a_{4},$ then $b_{1}, b_{2}, b_{3}$ and $b_{4}$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of these

Answer 1

Answer: (D)

Here, $b_{1}=a_{1}$
$b_{2}=a_{1}+a_{2}=a_{1}(1+r)$
$b_{3}=a_{1}(1+r)+a_{1} r^{2}=a_{1}\left(1+r+r^{2}\right)$
$b_{4}=a_{1}\left(1+r+r^{2}\right)+a_{1} r^{3}=a_{1}\left(1+r+r^{2}+r^{3}\right), r$ being the common ratio of the $G.P$.
Clearly, $b_{1}, b_{2}, b_{3}$ and $b_{4}$ are in neither of $A.P., G.P$. and $H.P$.